Linear Programming NCERT Solutions | Mathematics (Maths) Class 12 - JEE PDF Download

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NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
Exercise 12.1                                                                        page no: 513 
1. Maximise Z = 3x + 4y 
Subject to the constraints: 
Solution: 
The feasible region determined by the constraints, x + y = 4, x = 0, y = 0, is given below. 
 
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given 
below. 
Corner point Z = 3x + 4y  
O (0, 0) 0  
A (4, 0) 12  
B (0, 4) 16 Maximum 
Hence, the maximum value of Z is 16 at the point B (0, 4). 
2. Minimise Z = -3x + 4y 
subject to . 
Solution: 
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NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
Exercise 12.1                                                                        page no: 513 
1. Maximise Z = 3x + 4y 
Subject to the constraints: 
Solution: 
The feasible region determined by the constraints, x + y = 4, x = 0, y = 0, is given below. 
 
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given 
below. 
Corner point Z = 3x + 4y  
O (0, 0) 0  
A (4, 0) 12  
B (0, 4) 16 Maximum 
Hence, the maximum value of Z is 16 at the point B (0, 4). 
2. Minimise Z = -3x + 4y 
subject to . 
Solution: 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
The feasible region determined by the system of constraints, 
is given below. 
 
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region. 
The values of Z at these corner points are given below. 
Corner point Z = – 3x + 4y  
O (0, 0) 0  
A (4, 0) -12 Minimum 
B (2, 3) 6  
C (0, 4) 16  
Hence, the minimum value of Z is – 12 at the point (4, 0). 
3. Maximise Z = 5x + 3y 
subject to . 
Solution: 
The feasible region determined by the system of constraints, 3x + 5y = 15, 5x + 2y = 10, x = 0, and y = 0, is given 
below. 
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NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
Exercise 12.1                                                                        page no: 513 
1. Maximise Z = 3x + 4y 
Subject to the constraints: 
Solution: 
The feasible region determined by the constraints, x + y = 4, x = 0, y = 0, is given below. 
 
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given 
below. 
Corner point Z = 3x + 4y  
O (0, 0) 0  
A (4, 0) 12  
B (0, 4) 16 Maximum 
Hence, the maximum value of Z is 16 at the point B (0, 4). 
2. Minimise Z = -3x + 4y 
subject to . 
Solution: 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
The feasible region determined by the system of constraints, 
is given below. 
 
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region. 
The values of Z at these corner points are given below. 
Corner point Z = – 3x + 4y  
O (0, 0) 0  
A (4, 0) -12 Minimum 
B (2, 3) 6  
C (0, 4) 16  
Hence, the minimum value of Z is – 12 at the point (4, 0). 
3. Maximise Z = 5x + 3y 
subject to . 
Solution: 
The feasible region determined by the system of constraints, 3x + 5y = 15, 5x + 2y = 10, x = 0, and y = 0, is given 
below. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
 
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these 
corner points are given below. 
Corner point Z = 5x + 3y  
O (0, 0) 0  
A (2, 0) 10  
B (0, 3) 9  
C (20 / 19, 45 / 19) 235 / 19 Maximum 
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19). 
4. Minimise Z = 3x + 5y 
such that . 
Solution: 
The feasible region determined by the system of constraints, x + 3y = 3, x + y = 2, and x, y = 0, is given below. 
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NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
Exercise 12.1                                                                        page no: 513 
1. Maximise Z = 3x + 4y 
Subject to the constraints: 
Solution: 
The feasible region determined by the constraints, x + y = 4, x = 0, y = 0, is given below. 
 
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given 
below. 
Corner point Z = 3x + 4y  
O (0, 0) 0  
A (4, 0) 12  
B (0, 4) 16 Maximum 
Hence, the maximum value of Z is 16 at the point B (0, 4). 
2. Minimise Z = -3x + 4y 
subject to . 
Solution: 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
The feasible region determined by the system of constraints, 
is given below. 
 
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region. 
The values of Z at these corner points are given below. 
Corner point Z = – 3x + 4y  
O (0, 0) 0  
A (4, 0) -12 Minimum 
B (2, 3) 6  
C (0, 4) 16  
Hence, the minimum value of Z is – 12 at the point (4, 0). 
3. Maximise Z = 5x + 3y 
subject to . 
Solution: 
The feasible region determined by the system of constraints, 3x + 5y = 15, 5x + 2y = 10, x = 0, and y = 0, is given 
below. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
 
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these 
corner points are given below. 
Corner point Z = 5x + 3y  
O (0, 0) 0  
A (2, 0) 10  
B (0, 3) 9  
C (20 / 19, 45 / 19) 235 / 19 Maximum 
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19). 
4. Minimise Z = 3x + 5y 
such that . 
Solution: 
The feasible region determined by the system of constraints, x + 3y = 3, x + y = 2, and x, y = 0, is given below. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
 
It can be seen that the feasible region is unbounded. 
The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2). 
The values of Z at these corner points are given below. 
Corner point Z = 3x + 5y  
A (3, 0) 9  
B (3 / 2, 1 / 2) 7 Smallest 
C (0, 2) 10  
7 may or may not be the minimum value of Z because the feasible region is unbounded. 
For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half-plane has 
common points with the feasible region or not. 
Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7. 
Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2). 
5. Maximise Z = 3x + 2y 
subject to . 
Solution: 
Page 5


 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
Exercise 12.1                                                                        page no: 513 
1. Maximise Z = 3x + 4y 
Subject to the constraints: 
Solution: 
The feasible region determined by the constraints, x + y = 4, x = 0, y = 0, is given below. 
 
O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given 
below. 
Corner point Z = 3x + 4y  
O (0, 0) 0  
A (4, 0) 12  
B (0, 4) 16 Maximum 
Hence, the maximum value of Z is 16 at the point B (0, 4). 
2. Minimise Z = -3x + 4y 
subject to . 
Solution: 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
The feasible region determined by the system of constraints, 
is given below. 
 
O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region. 
The values of Z at these corner points are given below. 
Corner point Z = – 3x + 4y  
O (0, 0) 0  
A (4, 0) -12 Minimum 
B (2, 3) 6  
C (0, 4) 16  
Hence, the minimum value of Z is – 12 at the point (4, 0). 
3. Maximise Z = 5x + 3y 
subject to . 
Solution: 
The feasible region determined by the system of constraints, 3x + 5y = 15, 5x + 2y = 10, x = 0, and y = 0, is given 
below. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
 
O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these 
corner points are given below. 
Corner point Z = 5x + 3y  
O (0, 0) 0  
A (2, 0) 10  
B (0, 3) 9  
C (20 / 19, 45 / 19) 235 / 19 Maximum 
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19). 
4. Minimise Z = 3x + 5y 
such that . 
Solution: 
The feasible region determined by the system of constraints, x + 3y = 3, x + y = 2, and x, y = 0, is given below. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
 
It can be seen that the feasible region is unbounded. 
The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2). 
The values of Z at these corner points are given below. 
Corner point Z = 3x + 5y  
A (3, 0) 9  
B (3 / 2, 1 / 2) 7 Smallest 
C (0, 2) 10  
7 may or may not be the minimum value of Z because the feasible region is unbounded. 
For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half-plane has 
common points with the feasible region or not. 
Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7. 
Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2). 
5. Maximise Z = 3x + 2y 
subject to . 
Solution: 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 12 – 
Linear Programming  
The feasible region determined by the constraints, x + 2y = 10, 3x + y = 15, x = 0, and y = 0, is given below. 
 
A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the feasible region. 
The values of Z at these corner points are given below. 
Corner point Z = 3x + 2y  
A (5, 0) 15  
B (4, 3) 18 Maximum 
C (0, 5) 10  
Hence, the maximum value of Z is 18 at points (4, 3). 
6. Minimise Z = x + 2y 
subject to 
. 
Solution: 
The feasible region determined by the constraints, 2x + y = 3, x + 2y = 6, x = 0, and y = 0, is given below.